Towers of Hanoi: Difference between revisions

=={{header|TI-83 BASIC}}==

TI-83 BASIC lacks recursion, so technically this task is impossible, however here is a version that uses an iterative method.

<lang ti-83b>PROGRAM:TOHSOLVE

0→A

1→B

0→C

0→D

0→M

1→R

While A<1 or A>7

Input "No. of rings=?",A

End

randM(A+1,3)→[C]

[[1,2][1,3][2,3]]→[E]

Fill(0,[C])

For(I,1,A,1)

I?[C](I,1)

End

ClrHome

While [C](1,3)≠1 and [C](1,2)≠1

For(J,1,3)

For(I,1,A)

If [C](I,J)≠0:Then

Output(I+1,3J,[C](I,J))

End

End

End

While C=0

Output(1,3B," ")

1→I

[E](R,2)→J

While [C](I,J)=0 and I≤A

I+1→I

End

[C](I,J)→D

1→I

[E](R,1)→J

While [C](I,J)=0 and I≤A

I+1→I

End

If (D<[C](I,J) and D≠0) or [C](I,J)=0:Then

[E](R,2)→B

Else

[E](R,1)→B

End

1→I

While [C](I,B)=0 and I≤A

I+1→I

End

If I≤A:Then

[C](I,B)→C

0→[C](I,B)

Output(I+1,3B," ")

End

Output(1,3B,"V")

End

While C≠0

Output(1,3B," ")

If B=[E](R,2):Then

[E](R,1)→B

Else

[E](R,2)→B

End

1→I

While [C](I,B)=0 and I≤A

I+1→I

End

If [C](I,B)=0 or [C](I,B)>C:Then

C→[C](I-1,B)

0→C

M+1→M

End

End

Output(1,3B,"V")

R+1→R

If R=4:Then:1→R:End

End

</lang>